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McGraw Hill Glencoe Algebra 2, 2012 View details

5. Law of Cosines

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Exercise 5 Page 825

Begin by using the Law of Sines if two angles and any side are given, or if two sides and an angle opposite one of them are given. Begin by using the Law of Cosines if two sides and their included angle are given, or if the three sides are given.

Law: Sines
Missing Parts: B≈ 40^(∘), C≈ 33^(∘), c≈ 6.8

Practice makes perfect

We can use the Law of Sines and the Law of Cosines to find missing measurements in triangles. Finding all of the missing measurements is called solving the triangle. To do this, we have to determine which of the two laws must be used first.

Solving Triangles
Given	Begin by Using
Two angles and any sides	Law of Sines
Two sides and an angle opposite one of them	Law of Sines
Two sides and their included angle	Law of Cosines
Three sides	Law of Cosines

Let's determine which law to use first and then solve the triangle.

Determining the Law

Let's begin by color coding the opposite angles and sides in the given triangle.

We have information about two sides and the angle opposite one of them. Therefore, we must start solving the triangle by using the Law of Sines.

Solving the Triangle

We will find the missing side length and the missing angle measures one at a time.

Finding the Angle Measures

Let's consider the Law of Sines and adapt it to our triangle. sin A/a = sin B/b We can substitute the given values into this equation and solve for B.

sin A/a = sin B/b

SubstituteValues

Substitute values

sin 107^(∘)/12 = sin B/8

▼

Solve for B

MultEqn

LHS * 8=RHS* 8

sin 107^(∘)/12* 8 = sin B

MoveRightFacToNum

a/c* b = a* b/c

sin 107^(∘) * 8/12 = sin B

ReduceFrac

a/b=.a /4./.b /4.

sin 107^(∘) * 2/3 = sin B

RearrangeEqn

Rearrange equation

sin B = sin 107^(∘) * 2/3

sin ^(- 1)(LHS)=sin ^(- 1)(RHS)

B = sin ^(- 1)(sin 107^(∘) * 2/3)

UseCalc

Use a calculator

B = 39.608367... ^(∘)

RoundInt

Round to nearest integer

B ≈ 40^(∘)

Finally, we can use the Triangle Angle Sum Theorem to find the measure of C. 107^(∘)+ 40^(∘)+ C=180^(∘) ⇕ C= 33^(∘)

Finding the Side Length

Let's add the newly obtained information to our diagram.

Since now we have three angle measures and two side lengths, we can use the Law of Sines. sin C/c=sin A/a Let's substitute C= 33^(∘), A= 107^(∘), and a= 12 into the above equation and solve for c.

sin C/c=sin A/a

SubstituteValues

Substitute values

sin 33^(∘)/c=sin 107^(∘)/12

▼

Solve for c

MultEqn

LHS * c=RHS* c

sin 33^(∘) = sin 107^(∘)/12* c

MultEqn

LHS * 12/sin 107^(∘)=RHS* 12/sin 107^(∘)

sin 33^(∘) * 12/sin 107^(∘) = c